Question 18

If n is any odd number greater than 1, then $$n(n^2 - 1)$$ is

Solution

Let's say n=2k+1 where k=1,2,3....
So $$n(n^2-1) = 4k(k+1)(2k+1)$$
As above expression is already divisible by 4 simultaneously $$k(k+1)(2k+1)$$ will always be divisible by 6.
Hence complete term will be divisible by 24.

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If n is any odd number greater than 1, then $$n(n^2 - 1)$$ is

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